$解：(2)AC+AD为定值，理由如下：$
$由题意得AB//OP//O'P'$
$∵AB//OP$
$∴△ABC∽△OPC$
$∴\frac {AB}{OP}=\frac {AC}{OC}$
$∵AB=h，OP=O'P'=l，OA=a$
$∴\frac h{l}=\frac {AC}{a+AC}$
$∴AC=\frac {ah}{l-h}$
$同理可得AD=\frac {(m-a)h}{l-h}$
$∴AC+AD=\frac {mh}{l-h}$
$∴AC+AD为定值$
$(3)设点A到点O的距离为S_{1}，点A到影子顶端C的距离为S_{2}$
$∵AB//OP$
$∴△ABC∽△OPC$
$∴\frac {AB}{OP}=\frac {AC}{OC}$
$∵AB=h，OP=l，AC=S_{2}，OC=OA+AC=S_{1}+S_{2}$
$∴\frac h{l}=\frac {S_{2}}{S_{1}+S_{2}}$
$∴\frac l{h}-1=\frac {S_{1}}{S_{2}}$
$∴\frac {S_{1}}{S_{2}}=\frac {v_{1}}{v_{2}}=\frac {l-h}h$
$∴v_{2}=\frac {hv_{1}}{l-h}$
